When does $ 9k^2 (x-5)^2 - 125k^2 \geq (9+5k^2)(x^2 - 10x) + 225 $ have a unique solution

Question

$$ 9k^2 (x-5)^2 - 125k^2 \geq (9+5k^2)(x^2 - 10x) + 225 $$ For which value of the constant k below will the inequality have a unique solution? choices are:

$1/2014, 3/2, -9, 2014$

I have already simplified the equation to:

$(4k^2-9)(x^2-10x+25) >= 0 $

but I'm not sure how this will have a unique solution, I was tempted to answer $3/2$ but wont the inequality be always true, thus it wont have a solution? (since $0>=0$) I'm not sure what "a unique solution" means here anymore

Answer 1

I think it means a unique solution. I would be considering the possible values of $x^2-10x+25$ to identify the condition for a unique solution.

Answer 2

Note that

$(4k^2-9)(x^2-10x+25) = (4k^2-9)(x-5)^2$

and of course $(x-5)^2 \geq 0$. So if the other factor is negative ... well I'll let you take it from there.

Answer 3

This inequation reduces to $$(4k^2-9)(x-5)^2 \ge 0.$$ This implies that if $k<-3/2 ~or~k>3/2$ any real value of $x$ including $x=5$ is a solution of the givem inequation (many solutions)

But if $4k^2-9<0 \Rightarrow -3/2<k<3/2$, then $x=5$ is the only (unique) solution of the given equation (unique solution).

More interestingly if $k=\pm 3/2$ this equation again has any real value of $x$ including $x=5$ as a solution (many solutions).

Answer 4

The answer is $k=\frac{1}{2014}$.

In this case, the factor $4k^2-9<0$, and in order for the inequality

$$(4k^2-9)(x-5)^2 \ge 0$$

to hold, the following must holds

$$(x-5)^2=0$$

which yields the unique solution $x=5$. Note that other three choices of $k$ would allow multiple values of $x$.